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# Perfect Squares

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12 Jun 2009, 23:52
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Hi, I'm struggling to understand the concept behind a particular perfect squares question. Could anybody help?

If 375y = x^2, and x and y are positive integers then which of the following must be an integer?

I. y/15
II. y/30
III. y^2/25

As I'm new to this forum, please do advise if I've used incorrect symbols or haven't laid everything out as per standard practice. Thanks in advance.
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Re: Perfect Squares [#permalink]

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13 Jun 2009, 00:09
Welcome to GMAT Club!
There are easier ways to write the math symbols here - I have edited your expression below. You can learn how to post math symbols here: writing-mathematical-symbols-in-posts-72468.html. Usually, all you have to do is to highlight the text and hit the M button. Bu there are exceptions such as fractions.

If $$375y = x^2$$, and x and y are positive integers then which of the following must be an integer?

I. $$\frac{y}{15}$$

II. $$\frac{y}{30}$$

III. $$\frac{y^2}{25}$$

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Re: Perfect Squares [#permalink]

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13 Jun 2009, 03:07
if 375 * y = x ^2

x and y are integers u can write it as (5^3 * 3) * y = x^2

since x is an integer x ^2 is a perfect square . so to make LHS a perfect square y = 5 *3

so u can see y/15 and y^2/25 is an integer.

i would go for i and iii.
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Re: Perfect Squares [#permalink]

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13 Jun 2009, 03:56
3121gmat wrote:
Hi, I'm struggling to understand the concept behind a particular perfect squares question. Could anybody help?

If 375y = x^2, and x and y are positive integers then which of the following must be an integer?

I. y/15
II. y/30
III. y^2/25

As I'm new to this forum, please do advise if I've used incorrect symbols or haven't laid everything out as per standard practice. Thanks in advance.

Welcome on board!
1. Looking 375y=x^2 tells us that y has to also be 375 in order for x to be a perfect square.
2. Breakdown 375 to its prime factor---- (5*5*5*3)=y
3. Now it's much easier to compare each statement.
I. y/15----> Yes (quotient=25)
II.y/30----> No (quotient=25/2)
III.y^2/25----> Yes (quotient=5625)
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Re: Perfect Squares [#permalink]

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13 Jun 2009, 23:25
3121gmat wrote:
Hi, I'm struggling to understand the concept behind a particular perfect squares question. Could anybody help?

If 375y = x^2, and x and y are positive integers then which of the following must be an integer?

I. y/15
II. y/30
III. y^2/25

As I'm new to this forum, please do advise if I've used incorrect symbols or haven't laid everything out as per standard practice. Thanks in advance.

Just as 3121gmat, I'm having some difficulty in understanding this problem.

Here's my 2 cents:

y doesn't have to be 375. For example, if we had a statment like 9y = x^2, which will be valid if y=4 & x=6.

In this case, if y=15, x^2 = 375*15,
or x=75.

I will go with (I)
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Re: Perfect Squares [#permalink]

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14 Jun 2009, 10:50
Let me give my shot

$$375y=x^2$$
splitting the 375 = $$5^2*5*3$$
now as 375y is a square of positive integer and y is also a positive integer
implies that y has to be = 5*3*a^2, where a is an integer too

Why? let me show
375y=$$5^2*5*3*[5*3*a^2]$$ = $$25^2*3^2*a^2$$

$$a^2$$ is added so as to generale the possible values of y.
If you don't understand till here try different values of 'a' such as {2,3,4} to get y and x. It will give you different set of values to understand the reason of using a.

1) $$y/15 = 15*a^2/15 = a^2$$
definitely a integer that too positive

2) $$y/30 = 15*a^2/30 = a^2/2$$
not necessary an integer if a^2 is not a multiple of 2.

3) $$y^2/25 = 15*15*a^4/25 = 9a^4$$
definitely an integer.

(i) and (iii)
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Re: Perfect Squares [#permalink]

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14 Jun 2009, 10:56
3121gmat wrote:
Hi, I'm struggling to understand the concept behind a particular perfect squares question. Could anybody help?

If 375y = x^2, and x and y are positive integers then which of the following must be an integer?

I. y/15
II. y/30
III. y^2/25

As I'm new to this forum, please do advise if I've used incorrect symbols or haven't laid everything out as per standard practice. Thanks in advance.

375y= 3.5.5.5.y = x^2 ==> the number of 3 is even number, the number of 5 is even number ==> y contains both 3 and 5.

Clearly, (I) and (III) meet requirement.
Re: Perfect Squares   [#permalink] 14 Jun 2009, 10:56
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# Perfect Squares

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